Question

In the given figure, ∠BCA = 120º and AB = c, BC = a, AC = b. Then

(a) c² = a² + b² + ba 

(b) c² = a² + b² – ba 

(c) c² = a² + b² – 2ba 

(d) c² = a² + b² + 2ab

Answer 1

(a) c² = a² + b² + ba

By the cosine rule, we have, 

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

⇒ cos 120º = \(\frac{a^2+b^2-c^2}{2ab}\)

⇒ \(-\frac{1}{2}\) = \(\frac{a^2+b^2-c^2}{2ab}\)

⇒ c² = a² + b² + ba